euclidean-geometrygeometryproof-writingtriangles
I'm interested in the following problem:
$ABC$ is a triangle and the line $YCX$ is parallel to $AB$ such that $AX$ and $BY$ are the angular bisectors of angle $A$ and angle $B$ respectively. If $AX$ meets $BC$ at $D$ and $BY$ meets $AC$ at $E$ and if $YE =XD$, then prove that $AC=BC$.
My attempt. I tried using some angle chasing then I used cosine formula for finding the two equal segment but that goes too lengthy.
Any help will be highly appreciated!
The center of the circle $XAX'$ it's a mid-point of $XX'$ because $$\measuredangle XAX'=90^{\circ}.$$ Let $M$ be this mid-point, $O$ be circumcenter of $\Delta ABC$ and let $AX$ is placed between rays $AO$ and $AM$.
Thus, in the standard notaition: $$\measuredangle OAM=\measuredangle OAX+\measuredangle XAM=\frac{\alpha}{2}-\measuredangle BAO+\measuredangle AXM=$$ $$=\frac{\alpha}{2}-\left(90^{\circ}-\gamma\right)+\beta+\frac{\alpha}{2}=90^{\circ}$$ and we are done!
It's well known that $DE, IF, CM$ are concurrent. Call this point $X$.
Then $CX\perp IK$, $CI\perp XK$, meaning $I$ is the orthocenter of $CXK$ and the result follows.
Another interesting thing is that $\triangle CXK$ is self-polar wrt the incircle.
Edit:
The following proof that $DE,IF,CM$ are concurrent is not mine. It's taken from Evan Chen's EGMO.
Let $X$ be the intersection of $DE$ and $FI$ and let $A',B'$ be points on $CA$ and $CB$ such that $A'B'\parallel AB$ and $A'B'$ passes through $X$.
Now $E$ is the foot of $I$ onto $CA'$, $X$ is the foot of $I$ onto $A'B'$ ($B'A'\parallel BA$ and $IF\perp AB$, so $IF\perp AB$), and $D$ is the foot of $I$ onto $CB'$. Then $EXD$ is in fact the Simson line of $I$ wrt $\triangle CA'B'$, which means that $I$ is on the circumcircle of $\triangle CA'B'$.
Since $CI$ is an angle bisector of $\angle A'CB'$, $I$ is the arc midpoint of arc $B'A'$, and so $XI$ is the perpendicular bisector of $B'A'$. This means $X$ is the midpoint of $B'A'$, and so taking a homothety at $A$ which takes $A'B'\mapsto AB$ shows that the $C$-median passes through $X$ as well.
Once seeing $I$ is the orthocenter of $\triangle CXK$, $CK\perp IX$ but we already know $IX\perp AB$ so $CK\parallel AB$.
Best Answer
Since $XY||AB$, in the standard notation we obtain:
$\measuredangle CXD=\frac{\alpha}{2},$ $\measuredangle CYE=\frac{\beta}{2},$ $\measuredangle DCX=\beta$, $\measuredangle ECY=\alpha$, $CX=b$ and $CY=a.$
Thus, by the law of sines for $\Delta DCX$ we obtain: $$\frac{XD}{\sin\beta}=\frac{b}{\sin\left(\beta+\frac{\alpha}{2}\right)},$$ which gives $$XD=\frac{b\sin\beta}{\sin\left(\beta+\frac{\alpha}{2}\right)}=\frac{bl_a}{c}.$$ By the same way we can get: $$YE=\frac{al_b}{c},$$ which gives $$al_b=bl_a.$$ Now, $$l_a=\frac{2bc\cos\frac{\alpha}{2}}{b+c}=\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2}}{2}}}{b+c}=\frac{2\sqrt{(a+b+c)(b+c-a)bc}}{b+c}.$$ By the same way: $$l_b=\frac{2\sqrt{(a+b+c)(a+c-b)ac}}{a+c}.$$ Thus, $$a^3(a+c-b)(b+c)^2=b^3(b+c-a)(a+c)^2.$$ Now, let $a>b$.
Thus, $$a+c-b>b+c-a,$$ $$a^2(b+c)^2>b^2(a+c)^2,$$ which gives $$a^3(a+c-b)(b+c)^2>b^3(b+c-a)(a+c)^2,$$ which is a contradiction.
By the same way if $a<b$ so $$a^3(a+c-b)(b+c)^2<b^3(b+c-a)(a+c)^2,$$ which is a contradiction again.
Id est, $a=b$ and we are done!